Question

The number of ways in which a mixed double game can be arranged from amongst 5 married couples if no husband and wife play in the same game is:
A. 56
B. 60
C. 96
D. None of these

Answer 1

Hint: In this question, we have to use the concept of combination to find the number of ways in which a mixed double game can be formed from 5 married couples if no husband and wife participate in the same game.

Formula used: If we want choose r things out of n, and then combination aspect is used. Mathematically, it is expressed as \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Also, \[n! = n \times \left( {n - 1} \right) \times .... \times 1\]

Complete step-by-step solution:
Here, we need four people for a mixed double game.
That is, we require two men and two women.
Suppose the sides of the game are P and Q. We have been given that there are 5 husbands and 5 wives. Let us choose 2 husbands for two sides P and Q can be selected.
Thus, we get
\[
  {}^5{C_2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} \\
   \Rightarrow {}^5{C_2} = \dfrac{{5!}}{{2!\left( 3 \right)!}} \\
   \Rightarrow {}^5{C_2} = \dfrac{{5 \times 4 \times 3!}}{{2! \times \left( 3 \right)!}} \\
   \Rightarrow {}^5{C_2} = \dfrac{{5 \times 4}}{{2!}} \\
 \]
By simplifying further, we get
\[
   \Rightarrow {}^5{C_2} = \dfrac{{20}}{{2 \times 1}} \\
   \Rightarrow {}^5{C_2} = 10 \\
 \]
Following the selection of the two spouses, their wives are to be eliminated as no husband and wife participate in the same game.
So, we need to choose 2 wives from the remaining \[\left( {5 - 2} \right) = 3\] wives.
Thus, we get
\[
  {}^3{C_2} = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \\
   \Rightarrow {}^3{C_2} = \dfrac{{3!}}{{2! \times 1!}} \\
   \Rightarrow {}^3{C_2} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}} \\
   \Rightarrow {}^3{C_2} = 3 \\
 \]
Also, two wives can interchange their sides P and Q in \[2! = 2 \times 1 = 2\] ways.
According to multiplication rule, the total number of ways \[ = {}^5{C_2} \times {}^3{C_2} \times 2\]
Let us simplify this.
Number of ways \[ = 10 \times 3 \times 2 = 60\]
Hence, the number of ways in which a mixed double game can be arranged from amongst 5 married couples if no husband and wife play in the same game is 60.
Therefore, the correct option is (B).

Additional Information: A combination is a selection of all or part of a set of items, regardless of their sequence of selection.

Note: Many students make mistakes in the calculation part. They may get confused while selecting wives for the mixed double game for finding the result. This is the only way through which we can solve this example in an easy manner.